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What is #int_1^oo 1/(1+x )-1/xdx#?

Answer 1

Scarlett Allison

# = ln (1/2)#

#int_1^oo 1/(1+x )-1/xdx#

#= [ ln(1+x )- ln x ]_1^oo#

#= [ ln((1+x )/x) ]_1^oo#

#= [ ln((1/x+1 )/1) ]_1^oo#

#= lim_{x to oo} ln((1/x+1 )/1) - ln((1/1+1 )/1) #

#= ln 1 - ln 2 = ln (1/2)#

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Answer 2

Evelyn Agard

The integral ( \int_{1}^{\infty} \left(\frac{1}{1+x} - \frac{1}{x}\right)dx ) diverges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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